It depends on the current load. If you're only pulling low amperage, the voltage drop will be minimal. If you're attempting to pull anywhere near 20 amps, the voltage drop will be considerable. It also depends on the type and quality of the wire.
The voltage drop depends on the current through the cable.For DC current in cable of 16 mm diameter, at 68° F, the voltage drop is(0.00857) x (current, Amperes) volts.
For appliances that need a lot of power, the advantage of a higher voltage is that less current is needed, because power = voltage x current. That means that the cable doesn't need to be so thick. It also means that for a given percentage of voltage drop in the cable, the allowable volt-drop in the cable is higher when the supply voltage is higher. This also allows the use of a thinner cable especially when the cable is 100 ft long or more.
This is a voltage drop question and a voltage needs to be stated.
This is a voltage drop to establish wire size question. For a correct answer to this question two values are needed. One value needed is the voltage and the other is whether it is three phase or a single phase installation.
It depends on the voltage being used. That is because the size of the cable is determined by its resistance, and to calculate the allowable resistance you need to know the voltage drop. Lower resistance means a thicker cable. Normally the allowable voltage drop is a percentage of the supply voltage, 5% for example. On a 120 v system this would allow a 6 v drop while on a 240 v system the voltage drop could be 12 v. So for a given load current, the cable for a 120 v system would need half the resistance and double the cross-section area than a cable for 240 v. But for a given amount of power, the current on the 240 v system would be halved, so the cable resistance could be four times higher and its cross-section one quarter of that needed for 120 v. That is why higher voltages are used to transmit power over long distances.
It depends on the material of the cable (aluminum or copper) and the gauge of the cable. (Thickness). And on the current you intend it to carry.
Wattage is really what is needed. If you are working with DC voltage, Ohm's Law is at work. But to answer your question directly probably a #4 wire. <<>> To answer this question the circuit's voltage needs to be stated. Then a voltage drop calculation can be made.
This is a voltage drop question. A voltage must be stated to answer this question.
This is a voltage drop question. A voltage at 30 amps needs to be stated to answer the question.
This is a voltage drop question. To answer this question a voltage has to be stated. The higher the voltage to the circuit becomes the smaller the wire size needed. After a certain voltage point the wire size will remain constant and the voltage drop at the load will become smaller.
This is a voltage drop question. A #1 copper conductor will limit the voltage drop to 3% or less when supplying 100 amps for 200 feet on a 240 volt system. Or a 3/0 copper conductor will limit the voltage drop to 3% or less when supplying 100 amps for 200 feet on a 120 volt system. In your question you sis not stipulate what the working voltage is.
Voltage drop is caused by resistance. From the equation V=IR the voltage across that resistance can be calculated. For a uniform conductor the resistance is linear with the length R=kx where k is in ohms per foot. For a given current, the voltage drop V=Ikx so the voltage drop per foot is Ik. Voltage drop per foot can be measured, allowing a calculation of voltage drop for very long lengths of conductor such as power transmission lines. This is a reasonable approximation as long as the total voltage drop in transmission is small relative to the supply voltage. If the voltage drop is large, the current will be limited by the total resistance. This implies that long distance transmission lines should be high voltage, because 1 megawatt at 10,000 volts requires 100 amps, while at 100,000 volts it only requires 10 amps. The voltage drop per foot for the same conductor would be 10 times as large at 10 times the current, and the power loss (I*IR) would be 100 times as large. For alternating current, inductance can be a factor; this implies that relatively low frequency would be preferred.